Jaap's Puzzle Page

Alexander's Star

This puzzle is in the shape of a great dodecahedron, which is one of the
Kepler-Poinsot polyhedra. As a puzzle it has 12 vertices which are the axes
of rotation. There are 30 moving pieces which have two colours each. Each
piece forms the edge between two vertices, and five such edges join at each
vertex. The great dodecahedron has 12 pentagonal faces, but as these go
through the interior of the solid, only part of each pentagon is visible
from the outside. The visible part of a pentagon consists of a ring of five
triangles. Each ring must be all of one colour, and rings on opposite sides
of the puzzle must be the same colour.

This puzzle is equivalent to just the edges of the megaminx. There is
however one minor difficulty with the star because only six colours are
used and therefore each colour combination occurs twice.

It is not terribly difficult to solve, but this puzzle is infamously hard
to twist. It was invented by Adam Alexander, a US mathematician, who
patented it on 26 March 1985, US 4,506,891.

The number of positions:

There are 30 pieces, with 2 orientations each, giving a maximum of
30!·230 positions. This limit is not reached because:

only even permutations of pieces are possible (2)

only and even number of flipped pieces are possible (2)

there are 15 pairs of indistinguishable pieces (215/2 since even
number of swaps)

the orientation of the puzzle does not matter (60)

This leaves 29!·213 = 72,431,714,252,715,638,411,621,302,272,000,000
or 7.2·1034 positions. There are however 12 solved positions because there
are 24 ways of arranging 5 colours around a star if you discount rotations, and two
of those occur in each solution.

Notation:

Let F, R, D be three adjacent vertices (Front, Right and Down), in other
words, F, R, D are arranged clockwise at the vertices of a triangle of
edges. Let U (Up) be the other vertex adjacent to both F and R, and let L
(Left) be the other vertex adjacent to F and D. Clockwise 1/5 twists of a
vertex are denoted by the appropriate letter, anti-clockwise 1/5 twists
by F', R', L', D' or U'. In the solution below, the D vertex is usually
still unsolved.

Solution:

This puzzle is related to the Megaminx, as it
is equivalent to solving only the edges of that puzzle.

Solve everything except the five pieces around one last vertex.
For this only two sequences are necessary. Suppose we want to place the
correct piece between the F and R vertices. First find the piece on the
puzzle (remember there are two possible pieces). Use any moves that don't
destroy solved parts of the puzzle to place the piece somewhere on the D
vertex. Then twist D and use one of the following sequences:

DF->FR:

R'D R

DR->RF:

F D'F'

Make sure you use the correct one, otherwise the piece is put in upside
down. If a piece is upside down in position, simply use either sequence
to displace it and then place it correctly. These sequences only disturb
the D vertex edges and the FR edge but nothing else.

Position the edges of the final vertex.
Rotate D to place as many edges as possible in their correct places.
You might find that the edges are in an odd permutation, for example you may have
three pieces correct and two that need to be swapped. In this case you
will have to swap any two identical pieces elsewhere on the puzzle and try
solving it again.
In this and the next phase, the vertices adjacent to D are in clockwise
order: L, F, R, b, B.
Hold the puzzle so that you can use one of the following:

DR->DF->DL->DR:

R D R' D R D'D' R'

DR->DL->DF->DR:

R DD R' D' R D' R'

DR->DF->DB->DR:

R D R' DD R DD R'

DR->DB->DF->DR:

R D'D' R' D'D' R D' R'

DR-DB, DF-DL:

R D R' D R D' R' DD R DD R'

Orient the edges of the final vertex.
The edges are flipped in pairs. It is impossible to have only one edge
upside down. Two similar sequences are used for this. Hold the puzzle so
that the unsolved vertex is D, one edge to be flipped is DR, and the other
is either at DL or DF: